2 characteristics and 2 differences for spiral and elliptical galaxies

2 answers:

5 0

Spiral galaxies have three main components: a bulge, disk, and halo (see right). The bulge is a spherical structure found in the center of the galaxy. This feature mostly contains older stars. The disk is made up of dust, gas, and younger stars. The disk forms arm structures. Our Sun is located in an arm of our galaxy, the Milky Way. The halo of a galaxy is a loose, spherical structure located around the bulge and some of the disk. The halo contains old clusters of stars, known as globular clusters.

Elliptical galaxies are shaped like a spheriod, or elongated sphere. In the sky, where we can only see two of their three dimensions, these galaxies look like elliptical, or oval, shaped disks. The light is smooth, with the surface brightness decreasing as you go farther out from the center. Elliptical galaxies are given a classification that corresponds to their elongation from a perfect circle, otherwise known as their ellipticity. The larger the number, the more elliptical the galaxy is. So, for example a galaxy of classification of E0 appears to be perfectly circular, while a classification of E7 is very flattened. The elliptical scale varies from E0 to E7. Elliptical galaxies have no particular axis of rotation.

saul85 [17]3 years ago

4 0

Spiral galaxies are in a spiral like motion while eliptical galaxies are shaped like an egg. Spiral galaxies look like they have a core while eliptical galaxies dont.

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What is the total pressure exerted by the mixture of gases only products or products + reactants?

dedylja [7]

Total = products + reactants

3 0

3 years ago

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$